Power-log Combo

Problem

Part 1: Calculate the indefinite integral

$$\int x^p \ln(x) \,dx.$$

Part 2: For what values $p$ does the integral $\int_0^1 x^p \ln(x) \,dx$ converge?

Solution

Part 1

This family of functions is the product of a power function with the natural logarithm function. Therefore the method of integration by parts should be used.

According to the LIATE strategy, let $u = \ln(x)$ and $dv = x^p dx$. Then $du = \frac{1}{x}dx$ and $\frac{{x}^{p+1}}{p+1}$. Therefore,

$$\begin{align} \int x^p \ln(x) \,dx &= \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p+1}}{p+1} \frac{1}{x} \,dx \\[5 pt] &= \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p}}{p+1} \,dx \\[5 pt] &= \ln(x) \frac{{x}^{p+1}}{p+1} - \frac{{x}^{p+1}}{(p+1)^2} + C. \end{align}$$

Part 2

From the result of the indefinite integral, it is obvious that $p \ne -1$. Just look at the denominators. Now this means

$$\int_0^1 x^p \ln(x) \,dx = \ln(x) \frac{{x}^{p+1}}{p+1} \Biggr|_0^1 - \frac{{x}^{p+1}}{(p+1)^2} \Biggr|_0^1.$$

Notice that if $p < -1$, the power functions become reciprocal functions, which means the integral diverges when evaluating the lower bound. However if $p > -1$, the integral converges to the expression

$$\int_0^1 x^p \ln(x) \,dx = \frac{-1}{(p+1)^2}.$$

Therefore the integral $\int_0^1 x^p \ln(x) \,dx$ converges if $p > -1$.